Question

Let $\mathbb{R}$ denote the set of all real numbers. Define the function $f: \mathbb{R} \rightarrow \mathbb{R}$ by

$$f(x) = \begin{cases} 2-2x^2 - x^2\sin{\frac{1}{x}} \quad \text{if } x \neq 0, \\ 2 \quad \text{if } x = 0. \end{cases}$$

Then which one of the following statements is TRUE?

• A. The function $f$ is NOT differentiable at $x=0$
• B. There is a positive real number $\delta$, such that $f$ is a decreasing function on the interval $(0, \delta)$
• C. For any positive real number $\delta$, the function $f$ is NOT an increasing function on the interval $(-\delta, 0)$
• D. $x=0$ is a point of local minima of $f$

Answer 1

Answer: (C)

For any positive real number $\delta$, the function $f$ is NOT an increasing function on the interval $(-\delta, 0)$.

Answer 2

The function is defined as

$$f(x) = \begin{cases} 2-2x^2 - x^2\sin{\frac{1}{x}} \quad \text{if } x \neq 0, \\ 2 \quad \text{if } x = 0. \end{cases}$$

Option (A): The function $f$ is NOT differentiable at $x=0$.

We calculate the derivative at $x=0$ using the definition: $f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = \lim_{h \to 0} \frac{(2 - 2h^2 - h^2\sin{\frac{1}{h}}) - 2}{h} = \lim_{h \to 0} \frac{-2h^2 - h^2\sin{\frac{1}{h}}}{h} = \lim_{h \to 0} (-2h - h\sin{\frac{1}{h}})$.

Since $-1 \le \sin{\frac{1}{h}} \le 1$, we have $-|h| \le h\sin{\frac{1}{h}} \le |h|$. By the Squeeze Theorem, $\lim_{h \to 0} h\sin{\frac{1}{h}} = 0$.

Thus, $f'(0) = \lim_{h \to 0} (-2h) - \lim_{h \to 0} (h\sin{\frac{1}{h}}) = 0 - 0 = 0$.

The derivative at $x=0$ exists and is equal to 0. So, $f$ is differentiable at $x=0$. Option (A) is false.

Option (B): There is a positive real number $\delta$, such that $f$ is a decreasing function on the interval $(0, \delta)$.

For $x \neq 0$, the derivative is $f'(x) = \frac{d}{dx}(2-2x^2-x^2\sin{\frac{1}{x}}) = -4x - (2x\sin{\frac{1}{x}} + x^2\cos{\frac{1}{x}}(-\frac{1}{x^2})) = -4x - 2x\sin{\frac{1}{x}} + \cos{\frac{1}{x}}$.

For $f$ to be decreasing on $(0, \delta)$, we need $f'(x) < 0$ for all $x \in (0, \delta)$.

$f'(x) = \cos{\frac{1}{x}} - 4x - 2x\sin{\frac{1}{x}}$.

As $x \to 0^+$, the terms $-4x$ and $-2x\sin{\frac{1}{x}}$ approach 0. The term $\cos{\frac{1}{x}}$ oscillates between -1 and 1.

Consider values of $x$ where $\frac{1}{x} = 2n\pi$ for large positive integer $n$. Then $x = \frac{1}{2n\pi}$. As $n \to \infty$, $x \to 0^+$.

For these values of $x$, $\cos{\frac{1}{x}} = \cos(2n\pi) = 1$ and $\sin{\frac{1}{x}} = \sin(2n\pi) = 0$.

$f'(x) = 1 - 4x - 2x(0) = 1 - 4x$.

For any $\delta > 0$, we can choose $n$ large enough such that $x = \frac{1}{2n\pi} < \delta$. For such $x$, $f'(x) = 1 - \frac{4}{2n\pi} = 1 - \frac{2}{n\pi}$.

If $n$ is large enough, $1 - \frac{2}{n\pi}$ is positive and close to 1. For example, if $n \ge 1$, $1 - \frac{2}{n\pi} > 1 - \frac{2}{\pi} > 0$.

Since there are values of $x$ in any interval $(0, \delta)$ where $f'(x) > 0$, $f$ is not decreasing on any interval $(0, \delta)$. Option (B) is false.

Option (C): For any positive real number $\delta$, the function $f$ is NOT an increasing function on the interval $(-\delta, 0)$.

For $f$ to be increasing on $(-\delta, 0)$, we need $f'(x) > 0$ for all $x \in (-\delta, 0)$.

Let $x \in (-\delta, 0)$, so $x = -y$ for $y \in (0, \delta)$.

$f'(x) = f'(-y) = -4(-y) - 2(-y)\sin{\frac{1}{-y}} + \cos{\frac{1}{-y}} = 4y + 2y\sin(-\frac{1}{y}) + \cos(\frac{1}{y}) = 4y - 2y\sin(\frac{1}{y}) + \cos(\frac{1}{y})$.

$f'(-y) = \cos(\frac{1}{y}) + 4y - 2y\sin(\frac{1}{y})$.

As $y \to 0^+$, the terms $4y$ and $-2y\sin(\frac{1}{y})$ approach 0. The term $\cos(\frac{1}{y})$ oscillates between -1 and 1.

Consider values of $y$ where $\frac{1}{y} = (2n+1)\pi$ for large positive integer $n$. Then $y = \frac{1}{(2n+1)\pi}$. As $n \to \infty$, $y \to 0^+$.

For these values of $y$, $\cos(\frac{1}{y}) = \cos((2n+1)\pi) = -1$ and $\sin(\frac{1}{y}) = \sin((2n+1)\pi) = 0$.

$f'(-y) = -1 + 4y - 2y(0) = -1 + 4y$.

For any $\delta > 0$, we can choose $n$ large enough such that $y = \frac{1}{(2n+1)\pi} < \delta$. For this $y$, $x = -y \in (-\delta, 0)$.

The value of the derivative is $f'(x) = -1 + 4y = -1 + \frac{4}{(2n+1)\pi}$.

We can choose $n$ large enough such that $\frac{4}{(2n+1)\pi} \le 1$ (e.g., $2n+1 \ge 4/\pi \approx 1.27$, so $n \ge 1$). For such $n$, $f'(x) = -1 + \frac{4}{(2n+1)\pi} \le 0$.

Since there are values of $x$ in any interval $(-\delta, 0)$ where $f'(x) \le 0$, $f$ is NOT an increasing function on the interval $(-\delta, 0)$. Option (C) is true.

Option (D): $x=0$ is a point of local minima of $f$.

For $x=0$ to be a local minima, there must exist some $\delta > 0$ such that $f(x) \ge f(0)$ for all $x \in (-\delta, \delta)$.

$f(0) = 2$. We need $f(x) \ge 2$ for $x \in (-\delta, \delta), x \neq 0$.

$2 - 2x^2 - x^2\sin{\frac{1}{x}} \ge 2$

$-2x^2 - x^2\sin{\frac{1}{x}} \ge 0$

$-x^2(2 + \sin{\frac{1}{x}}) \ge 0$.

Since $x^2 \ge 0$, we need $-(2 + \sin{\frac{1}{x}}) \ge 0$, which means $2 + \sin{\frac{1}{x}} \le 0$.

However, $-1 \le \sin{\frac{1}{x}} \le 1$, so $1 \le 2 + \sin{\frac{1}{x}} \le 3$.

Thus, $2 + \sin{\frac{1}{x}}$ is always positive.

So, $-x^2(2 + \sin{\frac{1}{x}}) \le 0$ for $x \neq 0$.

This means $f(x) = 2 - 2x^2 - x^2\sin{\frac{1}{x}} \le 2 = f(0)$ for $x \neq 0$.

For $x \neq 0$, $f(x) = 2 - x^2(2 + \sin(1/x))$. Since $x^2 > 0$ and $2 + \sin(1/x) \ge 2-1=1 > 0$, we have $x^2(2 + \sin(1/x)) > 0$.

Thus, $f(x) = 2 - (\text{positive quantity}) < 2 = f(0)$ for all $x \neq 0$.

This means $f(x) < f(0)$ for all $x \neq 0$.

Therefore, $x=0$ is a point of strict local maxima, not local minima. Option (D) is false.

The only true statement is (C).